Semester 3. Lecture4.

Lecture 4. Electric field of charged conductors.

The energy of the electrostatic field.

Field near the conductor. Capacitance of conductors and capacitors. (Capacities of flat, cylindrical and spherical capacitors). Energy of a system of fixed charges. The energy of a charged conductor, capacitor. Electrostatic field energy density.

When introducing a conductor into the outer electric field the charges inside the conductor begin to move under the action of forces from the external field until equilibrium is reached. This leads to a redistribution of the electric charge inside the conductor. Regions of the conductor, previously electrically neutral, acquire an uncompensated electric charge. Consequently, an electric field appears (or, as they say, is induced) in the conductor

. The condition for the equilibrium of electric charges:

,

those. field strength inside the conductor:

Therefore, from equality we get inside the conductor. Therefore, this condition is also satisfied at the boundary of the conductor. Those. conductor surface is equipotential surface , that's why the lines of force of the electric field are perpendicular to the surface of the conductor at each point .

charged conductor .

If an external electric charge is imparted to a solitary conductor, then the condition for the equilibrium of charges again leads to the condition:

,inside the conductor.

It follows that all external charges are located on the surface of the conductor, since. the field strength inside the conductor is zero, and according to the Gauss theorem for any closed surface inside the conductor (including the outer surface of the conductor):

.

Since the surface of the conductor in this case is also equipotential, the lines of force of the electric field are directed perpendicular to the surface of the conductor at each of its points.

From the Gauss theorem it follows that near the surface of the conductor

The magnitude of the electric displacement vector is equal to the surface density of external charges.

The charge on the surface of the conductor is distributed in such a way that the surface potential remains constant. This leads to the fact that the charge density on the surface of the conductor is not the same. For example, on the sharp parts of the conductors, the charge density is greater than in the recesses. In this regard, various phenomena arise, for example, "charge drain". If the conductor is in the air, then ionization of the air occurs near the tip, carrying away part of the electric charge - a phenomenon called "electric wind".

Electrical Imaging Method .

If the equipotential surface is replaced by a conducting surface, and then the part of the field that this surface separates is discarded, then the field pattern in the remaining part will not change. Conversely, if the field picture is supplemented with fictitious charges so that the conducting surface can be replaced by an equipotential one, then the initial field picture will not change.

Example.Find the force of attraction of a point charge to an infinite conducting plane . To do this, we will supplement the picture with another charge of the same type, but of the opposite sign, located symmetrically with respect to the plane. Then the plane will coincide with the equipotential surface, so the plane can be discarded and the force of interaction between charges can be found: .

The energy of a charged conductor .

The energy of a solitary charged conductor is defined as the energy of a system of charges: . On the conductor, so the energy of a solitary conductor:

.

For a system of charged conductors: .

In particular, for two conductors having charges q of the same magnitude but different in sign, the energy will be equal to: .

Comment . The magnitude of the potential difference called tension between bodies.

Experience shows that there is a linear relationship between the charge of a solitary conductor and its potential: . Proportionality factor FROM called coefficient of electrical containers or electrical capacity . The unit of electrical capacity is Farad (

).

Capacitor is called a system of two conductors charged with the same magnitude, but different in sign charges. The conductors are called capacitor plates .

The capacitance of a capacitor is determined by the formula.

The capacitor is conventionally designated.

Connection of capacitors

Consider a series connection of two capacitors C 1 and C 2. Point A between the capacitors is separated from the rest of the circuit, so its electrical charge cannot change. Since the initial charge of any point was equal to zero, then . Consequently, the charges of the capacitor plates adjacent to point A are equal in magnitude, but opposite in sign. But since the value of the charge of the plates is equal to the charge of the capacitors, then. The total charge of point A is zero, so if we discard this point along with the plates, then nothing will change in the circuit. Because the charges of the extreme plates are also the same in magnitude, but different in sign, then the resulting capacitor will have the same charge in magnitude.

TOTAL . The charges of capacitors connected in series are the same in magnitude. The total charge of the capacitors connected in series is equal to the charge of each of the capacitors.

For this case, the total voltage is equal to the sum of the voltages on the capacitors: U GENERAL \u003d U 1 + U 2. The charges of the capacitors are the same: q 1 \u003d q 2 \u003d q. Then . That's why .

When capacitors are connected in series, their capacitances are added according to the law of reciprocals . 

Capacitance calculation for parallel connection of capacitors.

For this case, the voltages on the capacitors are the same: U 1 \u003d U 2 \u003d U.

The total charge is equal to the sum of the charges: q GEN = q 1 + q 2 or C GEN U=C 1 U+C 2 U.

Then C GENERAL =C 1 +C 2 . When capacitors are connected in parallel, their capacitances add up. 

Capacitor energy :

.

The total charge of the capacitor is zero. A capacitor stores electrical energy by separating electrical charges.

Examples for calculating the capacitance of capacitors .

Flat (air) condenser represents two parallel plates, the distance between which is much less than the dimensions of the plates, so that the field between the plates can be considered uniform. There is a vacuum (air) between the plates, therefore  = 1.

In this case, when calculating the field pattern, one can use the results obtained for the field of an infinite charged plane. Since the charges and areas of the plates are equal in magnitude, then the magnitude of the field strength created by each of the plates is the same: but the directions of the intensity vectors are different (the intensity vector from a negatively charged plate is shown by a dotted line). Between the plates, the intensity vectors are directed in the same way, so the total intensity is equal to the sum of the field strengths created by each of the plates:

.

Outside the plates, the field strength vectors are directed oppositely, so the field strength outside is zero. In this way, in a capacitor, the field strength is nonzero only between the plates.

Since the electrostatic field is a field of conservative force, the integral does not depend on the shape of the trajectory G, so the potential difference between the plates can be found along the perpendicular connecting the plates, the length of which is equal to d:, where d is the distance between the plates. Then the capacitance of a flat (air) capacitor in accordance with the definition will be equal to:

Cylindrical (air) condenser consists of two coaxial cylinders

of the same length, nested in each other so that the distance between the plates is much less than the dimensions of the plates.

Let the length of the capacitor L, the charge of the inner lining is positive: q > 0. Plating radii R 1 and R 2 , let R 1 <R 2. Field strength between the plates at a distance r from the inner lining, i.e. for R 1 <r <R 2 , we find using the Gauss theorem:

.

Then the voltage between the plates: .

Therefore, the electrical capacity of a cylindrical (air) capacitor: .

FROM spherical (air) condenser represents two nested concentric spheres with the radii of the plates R 1 and R 2 ,R 1 <R 2. Let the charge of the inner lining q> 0. The field strength between the linings at a distance r from the inner lining ( R 1 <r <R 2) we find by the Gauss theorem:

.

Tension between plates: .

Therefore, the capacitance of a spherical (air) capacitor .

Volumetric energy density of the electrostatic field.

Consider a flat air condenser. Energy of a charged capacitor

.

The amount of space between the plates of a capacitor. Since the field between the plates is considered to be homogeneous, the unit volume of this field has the energy . This value is called volumetric energy density .

In the case when the field is not uniform, the volumetric energy density is .

In matter, the volumetric energy density of the electric field .

In the case of a homogeneous isotropic dielectric, therefore .

Because , then , where

The energy of the electric field in vacuum is the energy of the polarization of matter.

Example . Consider a charged thin-walled sphere of radius R. Since charges of the same name repel each other on the sphere, the repulsive forces tend to stretch the surface of the sphere. We can assume that from inside the sphere, the walls are affected by additional pressure p, bursting the sphere and caused by the presence of an electric charge on the surface. Let's find R.

The field strength inside the sphere is zero, so the volume energy density of the electric field w is different from zero only outside the sphere.

With a slight increase in the radius of the sphere by dR its volume will increase, while in that part of the surrounding space that got inside the sphere, the volumetric energy density will become equal to zero. Therefore, the change in the energy of the field outside will be equal to, where S is the surface area. But with the expansion of the sphere, the pressure forces inside the sphere will do the work . Since then from where.

Example . Let's find the forces acting on the plates in a charged flat capacitor, disconnected from the power source.

The plates are oppositely charged, so they attract. Assume that the plates are close to each other by a small amount. x. Then the volume of the condenser is reduced by dV = xS, so the energy of the capacitor has decreased by dW = wdV. Attractive forces do work  A = fx. Since A= dW, then fx = wxS. Therefore, the magnitude of the force is F = wS. The additional pressure that these forces create is equal to.

The above examples show that bodies in an electric field are subject to forces that cause additional pressure equal to the volumetric energy density.

The pressure caused by the presence of an electric field is equal to the volumetric energy density .

Forces , acting on the body from the side of some field, are called pondemotor .

The oppositely charged capacitor plates attract each other.

Mechanical forces acting on macroscopic charged bodies are calledponderomotive .

We calculate the ponderomotive forces acting on the plates of a flat capacitor. In this case, two options are possible:

The capacitor is charged and disconnected from the charged battery(in this case, the number of charges on the plates remains constant q = const).

When one plate of a capacitor is removed from the other, work is done

due to which the potential energy of the system increases:

In this case, dA = dW . Equating the right sides of these expressions, we obtain

(12.67)

In this case, when differentiating, the distance between the plates was designated x.

Capacitor charged but not disconnected from battery(in this case, when moving one of the capacitor plates, the voltage will remain constant ( U = const). In this case, when one plate moves away from the other, the potential energy of the capacitor field decreases, since charges “leak” from the plates, therefore

But

, then

The resulting expression coincides with the formula

. It can also be represented in another form if instead of the charge q we introduce the surface density:

(12.68)

The field is uniform. The field strength of the capacitor is

, where x is the distance between the plates. Substituting into the formula

U 2 \u003d E 2 x 2, we get that the force of attraction of the plates of a flat capacitor

(12.69)

These forces act not only on the plates. Since the plates, in turn, put pressure on the dielectric placed between them and deform it, pressure arises in the dielectric

(S is the area of ​​each plate).

The pressure arising in the dielectric is

(12.70)

Examples of problem solving

Example 12.5. To flat plates air condenser a potential difference of 1.5 kV is applied. Plate area 150cm 2 and the distance between them is 5 mm. After disconnecting the capacitor from the voltage source, glass was inserted into the space between the plates (ε 2 =7). Define:

1) potential difference between the plates after the introduction of a dielectric; 2) the capacitance of the capacitor before and after the introduction of the dielectric; 3) the surface charge density on the plates before and after the introduction of the dielectric.

Given: U 1 \u003d 1.5 kV \u003d 1.5 ∙ 10 3 V; S \u003d 150cm 2 \u003d 1.5 ∙ 10 -2 m 2; ε 1 =1; d=5mm=5∙10 -3 m.

Find: 1) U 2 ; 2) C 1 C 2; 3) σ 1 , σ 2

Solution . Because

(σ is the surface charge density on the capacitor plates), then before the introduction of the dielectric σd \u003d U 1 ε 0 ε 1 and after the introduction of the dielectric σd \u003d U 2 ε 0 ε 2, therefore

The capacitance of the capacitor before and after the introduction of a dielectric

and

The charge of the plates after disconnection from the voltage source does not change, i.e. q=const. Therefore, the surface charge density on the plates before and after the introduction of the dielectric

Answer: 1) U 2 \u003d 214V; 2) C 1 \u003d 26.5 pF; C 2 \u003d 186pF; 3) σ 1 = σ 2 = 2.65 μC/m 2.

Example 12.7. The gap between the plates of a flat capacitor is filled with an anisotropic dielectric, the permeability ε of which varies in the direction perpendicular to the plates according to the linear lawε = α + βх from ε 1 up to ε 2 , and ε 2 > ε 1 . The area of ​​each liningS, the distance between themd. Find the capacitance of the capacitor.

Given:S; d; ε 1 ; ε 2

Find: FROM.

Solution . The dielectric constant ε varies linearly, ε = α + βx, where x is measured from the lining, whose permeability is equal to ε 1 . Considering that ε (0) = ε 1 , ε (d) = ε 2 , we obtain the dependence

. Find the potential difference between the plates:

The capacitance of the capacitor will be

Answer:

Example 12.7. Between the plates of a flat capacitor charged to a potential difference U , two layers of dielectrics are placed parallel to its plates. The thickness of the layers and the permittivity of the dielectrics are, respectively,d 1 , d 2 , ε 1 , ε 2 . Determine the strength of electrostatic fields in dielectric layers.

Given: U; d 1 , d 2 , ε 1 , ε 2

Find: E 1 , E 2 .

Solution . The voltage across the capacitor plates, given that the field within each of the dielectric layers is uniform,

U=E 1 d 1 +E 2 d 2 . (one)

The electrical displacement in both dielectric layers is the same, so we can write

D=D1=D2= ε 0 ε 1 E 1 = ε 0 ε 2 E 2 (2)

From expressions (1) and (2) we find the desired

(3)

From formula (2) it follows that

Answer:

;

Example 12.7. Plate area S flat capacitor is 100cm 2 . The space between the plates is filled closely with two layers of dielectrics - a mica plate (ε 1 =7) thick d 1 =3.5 mm and paraffin (ε 2 =2) thickness d 2 =5 mm. Determine the capacitance of this capacitor.

Given: S=100cm 2 =10 -2 m 2 ; ε 1 =7; d 1 =3.5mm=3.5∙10 -3 m;, ε 1 =2; d 1 =3.5mm=5∙10 -3 m;

Find: FROM.

Solution . Capacitor capacity

where = - charge on the capacitor plates (- surface charge density on the plates); \u003d - potential difference of the plates, equal to the sum of the voltages on the dielectric layers: U \u003d U 1 +U 2. Then

(1)

The voltages U 1 and U 2 are found by the formulas

;

(2)

where E 1 and E 2 - the strength of the electrostatic field in the first and second layers of the dielectric; D is the electrical displacement in dielectrics (the same in both cases). Taking into account that

And given formula (2), from expression (1) we find the desired capacitance of the capacitor

Answer: C \u003d 29.5pF.

Example 12.7. A battery of three capacitors connected in series C 1 \u003d 1 μF; FROM 2 \u003d 2uF and C 3 \u003d 4 μF are connected to an EMF source. Capacitor battery charge q \u003d 40 μC. Determine: 1) voltage U 1 , U 2 and U 3 on each capacitor; 2) EMF source; 3) the capacity of the capacitor bank.

Given : C 1 \u003d 1 μF \u003d 1 ∙ 10 -6 F; C 2 \u003d 2 μF \u003d 2 ∙ 10 -6 F and C 3 \u003d 4 μF \u003d 4 ∙ 10 -6 F; q \u003d 40 μC \u003d 40 ∙ 10 -6 F .

Find: 1) U 1 , U 2 , U 3 ; 2) ξ; 3) C.

Solution . When capacitors are connected in series, the charges of all plates are equal in absolute value, therefore

q 1 \u003d q 2 \u003d q 3 \u003d q.

Capacitor voltage

The EMF of the source is equal to the sum of the voltages of each of the series-connected capacitors:

ξ \u003d U 1 + U 2 + U 3

When connected in series, the reciprocals of the capacitances of each of the capacitors are summed up:

Where is the desired capacity of the capacitor bank

Answer: 1) U 1 \u003d 40V; U 2 \u003d 20V, U 3 = 10V; 2) Ɛ= 70V; 3) C \u003d 0.571 μF.

Example 12.7. Two flat air capacitors of the same capacity are connected in series and connected to an EMF source. How and how many times will the charge of capacitors change if one of them is immersed in oil with a dielectric constant ε=2.2.

Given: C 1 \u003d C 2 \u003d C; q \u003d 40 μC \u003d 40 ∙ 10 -6 F ; ε 1 =1; ε 2 =2,2.

Find: .

Solution . When capacitors are connected in series, the charges of both capacitors are equal in magnitude. Before immersion in a dielectric (in oil), the charge of each capacitor

where ξ \u003d U 1 + U 2 (when capacitors are connected in series, the EMF of the source is equal to the sum of the voltages of each of the capacitors).

After one of the capacitors is immersed in a dielectric, the charges of the capacitors are again the same and, respectively, on the first and second capacitors are equal

q= CU 1 =ε 2 CU 2

(taking into account that ε 1 =1), whence, if we take into account that ξ = U 1 + U 2 , we find

(2)

Dividing (2) by (1), we find the desired ratio

Answer:

, i.e. the charge of the capacitors increases by a factor of 1.37.

Example 12.7. Capacitors with capacitances C each are connected as shown in fig.a. determine the capacitance common this connection of capacitors. .

Solution . If you disconnect capacitor C 4 from the circuit, you get a connection of capacitors, which is easily calculated. Since the capacities of all capacitors are the same (C 2 \u003d C 3 and C 5 \u003d C 6), both parallel branches are symmetrical, therefore the potentials of points A and B, equally located in the branches, must be equal. Capacitor C 4 is thus connected to points with zero potential difference. Therefore, the capacitor C 4 is not charged, i.e. it can be excluded and the scheme presented in the condition of the problem can be simplified (Fig. b).

This circuit consists of three parallel branches, two of which contain two capacitors in series.

Answer: C total = 2C.

Example 12.7. Flat air condenser with capacity C 1 \u003d 4pF charged to a potential differenceU 1 =100V. After disconnecting the capacitor from the voltage source, the distance between the capacitor plates was doubled. Determine: 1) potential differenceU 2 on the capacitor plates after their separation; 2) the work of external forces to push the plates apart.

Given: C 1 \u003d 4pF \u003d 4 ∙ 10 -12 F; U 1 \u003d 100V; d 2 \u003d 2d 1.

Find: 1) U 2 ;2)A.

Solution . The charge of the capacitor plates after disconnection from the voltage source does not change, i.e. Q=const. That's why

C 1 U 1 \u003d C 2 U 2, (1)

where C 2 and U 2 are, respectively, the capacitance and the potential difference on the capacitor plates after they are moved apart.

Given that the capacitance of a flat capacitor

, from formula (1) we obtain the desired potential difference

(2)

After disconnecting the capacitor from the voltage source, the system of two charged plates can be considered as closed, for which the law of conservation of energy is satisfied: the work A of external forces is equal to the change in the energy of the system

A \u003d W 2 - W 1 (3)

where W 1 and W 2 are the energy of the capacitor field in the initial and final states, respectively.

Given that

and

(q – const), from formula (3) we obtain the desired work of external forces

[taken into account that q=C 1 U 1 and formula (2)].

Answer : 1) U 2 \u003d 200V; 2) A \u003d 40nJ.

Example 12.7. A solid ball of dielectric with a radiusR=5cm charged uniformly with bulk density ρ=5nC/m 3 . Determine the energy of the electrostatic field contained in the space surrounding the ball.

Given: R=5cm=5∙10 -2 m; ρ=5nC/m 3 = 5∙10 -9 C / m 3.

Find: W.

Solution . The field of a charged ball is spherically symmetrical, so the volumetric charge density is the same at all points located at equal distances from the center of the ball.

E energy in an elementary spherical layer (it is chosen outside the dielectric, where the energy should be determined) with a volume of dV (see figure)

where dV=4πr 2 dr (r is the radius of an elementary spherical layer; dr is its thickness);

(ε=1 – field in vacuum; E – electrostatic field intensity).

We will find the intensity E by the Gauss theorem for a field in vacuum, and mentally choose a sphere with radius r as a closed surface (see figure). In this case, the entire charge of the ball, which creates the field under consideration, gets inside the surface, and, according to the Gauss theorem,

Where

Substituting the found expressions into formula (1), we obtain

The energy contained in the space surrounding the ball,

Answer: W=6.16∙10 -13 J.

Example 12.7. Planar capacitor with the area of ​​the platesSand the distance between them ℓ the charge is reportedq, after which the capacitor is disconnected from the voltage source. Determine the force of attractionFbetween the capacitor plates, if the dielectric constant of the medium between the plates is equal to ε.

Given : S; ℓ; q; ε .

Find: F.

Solution . The charge of the capacitor plates after disconnection from the voltage source does not change, i.e. q=const. Suppose that under the action of the force of attraction F, the distance between the capacitor plates has changed by d ℓ . Then the force F does work

According to the law of conservation of energy, this work is equal to the energy loss of the capacitor, i.e.

. (3)

Substituting into the formula for the energy of a charged capacitor

expression for the capacitance of a flat capacitor

, we get

(4)

Answer:

Example 12.7. Flat plate capacitorSand the distance between them ℓ connected to a constant voltage sourceU. Determine the force of attractionFbetween the capacitor plates, if the dielectric constant of the medium between the plates is equal to ε.

Given : S; ℓ; U; ε .

Find: F.

Solution . According to the condition of the problem, a constant voltage is maintained on the capacitor plates, i.e. U=const. Suppose that under the action of the force of attraction F, the distance between the capacitor plates has changed by dℓ. Then the force F does work

According to the law of conservation of energy, this work in this case goes to increase the energy of the capacitor (compare with the previous task), i.e.

whence, based on expressions (1) and (2), we obtain

(3)

Substituting into the formula for the energy of the capacitor

expression for the capacitance of a flat capacitor

, we get

(4)

Substituting the energy value (4) into formula (3) and performing differentiation, we find the desired force of attraction between the capacitor plates

.

where the "-" sign indicates that the force F is an attractive force.

Answer :